Finding k for x is undefined at $x=2$

[karush]

If $f(x)=\begin{cases}
\dfrac{(2x+1)(x-2)}{x-2}, &\text{for } x\ne 2 \\[3pt] k, &\text{for } x=2 \\
\end{cases}$
for what number k will the function be continuous
a. 0 b. 1 c. 2 d. 3 e. 5
---------------------------------
I chose
[e] x is undefined at $x=2$ so
rewrite as $2x+1$
plug in $f(2)=2(2)+1=5=k$

hopefully
probably suggestions etc

 

[skeeter]

for f(x) to be continuous at x = 2 ...

$\displaystyle \lim_{x \to 2} f(x) = f(2)$

 

[karush]

So discontinuous. Means a hole always?

 

[skeeter]

So discontinuous. Means a hole always?

not just a hole (a removable discontinuity)

also, a vertical asymptote or a jump (non-removable discontinuities)